Question: Simplify; express your answer in exponential form. Assume $r\neq 0, n\neq 0$. $\dfrac{{(r^{-1})^{2}}}{{(r^{-4}n^{3})^{3}}}$
Solution: To start, try working on the numerator and the denominator independently. In the numerator, we have ${r^{-1}}$ to the exponent ${2}$ . Now ${-1 \times 2 = -2}$ , so ${(r^{-1})^{2} = r^{-2}}$ In the denominator, we can use the distributive property of exponents. ${(r^{-4}n^{3})^{3} = (r^{-4})^{3}(n^{3})^{3}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(r^{-1})^{2}}}{{(r^{-4}n^{3})^{3}}} = \dfrac{{r^{-2}}}{{r^{-12}n^{9}}}$ Break up the equation by variable and simplify. $\dfrac{{r^{-2}}}{{r^{-12}n^{9}}} = \dfrac{{r^{-2}}}{{r^{-12}}} \cdot \dfrac{{1}}{{n^{9}}} = r^{{-2} - {(-12)}} \cdot n^{- {9}} = r^{10}n^{-9}$.